Prove that \((x)\) is a maximal ideal in \(R[x]\). i.e. \((x) \subset I \subset R[x] \implies (x) = I \vee I = R[x]\).

Proof: Let us use the reductio ad absurdum technique, and assume that \((x) \neq I \wedge I \neq R[x]\). Then, \(f(x) \in I, f(x) \notin (x), f(x) \notin R[x]\).

Since \(f(x)\) is a general polynomial, we can use the remainder theorem to obtain: \[f(x) = g(x)q(x) + r(x)\]

Now, since we're free to choose \(g\), choose \(g(x) = x\). Then \(r(x)\) must be of a lower degree, and hence a constant \(c\). \[f(x) = x . q(x) + c\]

There are two cases to consider: either \(c = 0\) or \(c\) is a nonzero constant. In the first case, since \(f(x) = x . q(x)\), our assumption that \(f(x) \notin (x)\) breaks, and we have a contradiction. In the second, \(I\) would contain the unit ideal, and hence \(I = R[x]\), breaking the assumption that \(I \neq R[x]\). This leads to another contradiction, finishing off the proof \(\Box\)

Prove that \(x^2 + 1\) is a maximal ideal in \(R[x]\). i.e. \((x^2 + 1) \subset I \subset R[x] \implies (x^2 + 1) = I \vee I = R[x]\).

Proof: Let us again use proof by contradiction, and assume that \((x^2 + 1) \neq I \wedge I \neq R[x]\). Then, \(f(x) \in I, f(x) \notin (x^2 + 1), f(x) \notin R[x]\).

Since \(f(x)\) is a general polynomial, we can use the remainder theorem to obtain: \[f(x) = g(x)q(x) + r(x)\]

Now, choose \(g(x) = x + \alpha\); \(r(x)\) can only be a zero, because if it were a nonzero constant, \(f(x)\) would contain \(1\), and hence \(f(x) \in R[x]\) otherwise: \[f(x) = (x + \alpha)q(x)\]

Since \(f(x) \in I\) from our assumption, we get \(I = (x + \alpha)\). Note that \(x^2 + 1 \in I\), from our hypothesis: \[x^2 + 1 = (x + \alpha)h(x)\]

But \(x^2 + 1\) is irreducible in \(R[x]\); assume that it factors: \[x^2 + 1 = (x - a)(x - b) = x^2 - (a + b) . x + ab\] \[\implies a + b = 0 \wedge ab = 1 \implies a^2 = -1\]

This cannot happen in \(R[x]\), and we hence arrive at a contradiction \(\Box\)

Provide an example of an irreducible element which is not prime, in a ring of your choosing.

Solution: Consider the ring of Gauss integers \(\mathbb{Z}[\sqrt{-5}]\). Then \(6 = 2 . 3 = (1 + \sqrt{-5}) (1 + \sqrt{-5})\), and this ring is not a unique factorization domain. Let us inspect \(2\): it belongs to the multiplicative set containing \(2 . 3\), but it does not divide either \(1 + \sqrt{-5}\) or \(1 - \sqrt{-5}\).

Prime ideals are defined as \(A \backslash S\), where \(S\) is a multiplicative set such that: \[fg \in S \implies f \in S \wedge g \in S\]

Now, \(\mathbb{Z}[\sqrt{-5}]\) does not cleave into \(S\) and \(A \backslash S\) cleanly when \(6 \in S\). Since \(1 + \sqrt{-5}\) or \(1 - \sqrt{-5}\) are prime, \(2\) and \(3\) are disjoint from the set containing them, and so are in S. Hence, \(2\) is not prime, but is irreducible, as required \(\Box\)

Provide examples of prime ideals that are not maximal in \(\mathbb{Z}[x, y]\).

Solution: \((x), (y), (x + y)\) are prime ideals, but only the last one is maximal; the first two are irreducible.

Prove that every maximal ideal is a prime ideal in ring \(A\).

Proof: The definition of a prime ideal \(P\) is: \[fg \in P \implies f \in P \vee g \in P\]

The definition of a maximal ideal is: \[\not\exists I : M \subset I \subset A\]

From this definition, it should be clear that maximal ideals can never be the product of two different maximal ideals \(J\) and \(K\), for if it were, then \[J \subset JK \subset A \text{ or } K \subset JK \subset A\]

contradicting the definition of a maximal ideal. i.e. for \(j \in J, k \in K\) \[jk \in M \implies j \in M \vee k \in M\]

This is precisely the definition of a prime ideal \(\Box\)

Give an example of an algebraically closed ring, and an example of one that is not.

Solution: \(\mathbb{C}[x]\) is algebraically closed, because all polynomials in one variable in the ring have solutions in the ring. \(\mathbb{R}[x]\) is not algebraically closed, because the solution of, for instance, \(x^2 + 1 = 0\) does not lie in the ring \(\Box\)

List all the maximal ideals of \(\mathbb{C}[x]\) and \(\mathbb{R}[x]\).

Solution: The maximal ideals of \(\mathbb{C}[x]\) are \((x - \alpha)\), while those of \(\mathbb{R}[x]\) are \((x - a)\) and \((x - \alpha)(x - \bar{\alpha})\), where \(a\) is a real number, and \(\alpha\) is a complex number. Notice that all maximal ideals of \(\mathbb{C}\) can be factorized into monic factors, while in the case of \(\mathbb{R}\), quadratic factors are required \(\Box\)

Prove that \(P\) is a prime ideal in ring \(A\) iff \(A/P\) is an integral domain.

An integral domain is defined by \(pq = 0 \implies p = 0 \vee q = 0 \mid p, q \in A/P\).

Proof: Given that \(A/P\) is an integral domain, show that \(P\) is prime. Consider elements \(a, b \in A\). Elements of \(A/P\) are then of the form \(a + P, b + P\). Now, use the definition of an integral domain to write: \[(a + P)(b + P) = 0 + P \implies a + P = 0 + P \vee b + P = 0 + P\] \[ab + P = 0 + P \implies a - 0 \in P \vee b - 0 \in P\] \[ab \in P \implies a \in P \vee b \in P\]

This is exactly the definition of a prime ideal, as required.

Proof, the other way: To show \(A/P\) is an integral domain, given that \(P\) is a prime ideal. First, consider the surjective homomorphism \[A \xrightarrow[]{\varphi} A/P\]

From isomorphism theorem, \(\text{ker}(\varphi) = P = \{a \mid \varphi(a) = 0\}\). We know that: \[pq \in P \implies p \in P \vee q \in P\] \[\varphi(p)\varphi(q) \in (0) \implies \varphi(p) \in (0) \vee \varphi(q) \in (0)\]

This gives us the required condition for integral domain \(\Box\)

Prove that \(M\) is a maximal ideal in ring \(A\) iff \(A/M\) is a field.

A field is defined by \(pq = 0 \implies p = 0 \vee q = 0 \mid p, q \in A/M\) and \(\exists v^{-1}: v^{-1}v = vv^{-1} = 1 \mid v \in A/M\).

Proof: To show that \(M\) is maximal given that \(A/M\) is a field. For the zerodivisor proof, follow exactly. For the next part, assume \(w \in A\): \[(v + M)(w + M) = 1 + M\] \[vw - 1 \in M \implies vw \in 1 + M = (1)\]

hence \(vw\) is unit, which implies that \(w = v^{-1}\) as required.

Proof, the other way: To show that \(A/M\) is a field, given \(M\) is maximal. Consider the surjective homomorphism: \[A \xrightarrow[]{\varphi} A/M\]

Now, since \(A/M\) is a field, the only ideals are \((0)\) and \((1)\); by isomorphism theorem, there is a bijective correspondence between ideals of \(A\) containing \(M\) and those of \(A/M\). \[\varphi^{-1}(0) = ker(\varphi) = (0)\] \[\varphi^{-1}(1) = A \backslash (0) = M\]

Since \((0) \subset M \subset A\), we have shown that \(M\) is maximal, as required \(\Box\)

For ideals \(P\) and \(Q\), show that \(PQ\) is contained within \(P \cap Q\).

Proof: From isomorphism theorem, \[(P + Q)/Q \cong P/(P \cap Q)\] \[PQ/(P + Q) = P \cap Q\]

Since \(P + Q = \text{gcd}(P, Q)\), we see that unless the gcd is \(1\), \(PQ\) is strictly smaller than \(P \cap Q\) \(\Box\)